的Python：在多個列表

Question

我列出任意數量的列表中找到相同的物品，例如：

[[1,2,3], [3,4,5], [5,6,7], [7,8,9]] 

現在我想包含存在於多個列表中的所有元素的列表：

[3,5,7] 

我該怎麼做？

謝謝！

Answer 1

比較數字如你自己動手完成它：

seen = set() 
repeated = set() 
for l in list_of_lists: 
    for i in set(l): 
    if i in seen: 
     repeated.add(i) 
    else: 
     seen.add(i) 

順便說一句，這裏的一個襯墊（不計進口），一些人正在尋求（應該是比其他方法效率較低）

from itertools import * 
reduce(set.union, (starmap(set.intersection, combinations(map(set, ll), 2)))) 

Answer 2

扁平化，排序，1迴路用同樣的方法之前和之後

Answer 3

您可以使用一組看到http://docs.python.org/library/stdtypes.html#set

Answer 4

您可以使用字典來獲得每個

from collections import defaultdict 

init_list = [[1,2,3], [3,4,5], [5,6,7], [7,8,9]] 
#defaultdict, every new key will have a int(0) as default value 
d = defaultdict(int) 
for values in init_list: 
    #Transform each list in a set to avoid false positives like [[1,1],[2,2]] 
    for v in set(values): 
    d[v] += 1 

#Get only the ones that are more than once 
final_list = [ value for value,number in d.items() if number > 1 ] 

Answer 5

l=[[1,2,3], [3,4,5], [5,6,7], [7,8,9]] 
d={} 
for x in l: 
    for y in x: 
     if not d.has_key(y): 
      d[y]=0 
     d[y]+=1 
[x for x,y in d.iteritems() if y>1] 

Answer 6

引用的次數：http://docs.python.org/library/stdtypes.html#set

#!/usr/bin/python 

ll = [[1,2,3], [3,4,5], [5,6,7], [7,8,9]] 
ls = [set(l) for l in ll] 

su = ls[0] #union 
ssd = ls[0] #symmetric_difference 
for s in ls[1:]: 
    su = su.union(s) 
    ssd = ssd.symmetric_difference(s) 

result = su.difference(ssd) 
print list(result) 

=>

[3, 5, 7] 

修改，並採用FP，

ll = [[1,2,3], [3,4,5], [5,6,7], [7,8,9]] 

u = reduce(set.union, map(set, ll)) 
sd = reduce(set.symmetric_difference, map(set, ll)) 
print u - sd 

=>

[3, 5, 7] 

Answer 7

試試這個：

data = [[1,2,3], [3,4,5], [5,6,7], [7,8,9], [1,2,3]] 

res = set() 

for i in data: 
    for j in data: 
     if i is not j: 
      res |= set(i) & set(j) 

print res 

Answer 8

乾淨的方式很可能是使用減少：

def findCommon(L): 
    def R(a, b, seen=set()): 
     a.update(b & seen) 
     seen.update(b) 
     return a 
    return reduce(R, map(set, L), set()) 

result = findCommon([[1,2,3], [3,4,5], [5,6,7], [7,8,9]]) 

結果是一組，但只是做list(result)如果你真的需要一個列表。

Answer 9

>>> sets = [[1,2,3], [3,4,5], [5,6,7], [7,8,9]] 
>>> seen = set() 
>>> duplicates = set() 
>>> 
>>> for subset in map(set, sets) : 
...  duplicates |= (subset & seen) 
...  seen |= subset 
... 
>>> print(duplicates) 
set([3, 5, 7]) 
>>> 

我試圖用的map/reduce單行答案，但不能完全得到它。

Answer 10

這是我去：

seen = set() 
result = set() 
for s in map(set, [[1,2,3], [3,4,5], [5,6,7], [7,8,9]]): 
    result.update(s & seen) 
    seen.update(s) 
print result 

此打印：

set([3, 5, 7]) 

Answer 11

另一種簡單的解決方案（一行）：

set.intersection(*[set(list) for list in list_of_lists])